alleryIdeal gas equation:
- PV = mR
- PV = RT/n
- P + V = nRT
- PV = nRT
Show answer and solution
Correct answer: D. PV = nRT
PV = nRT.
n = moles, R = 8.314 J/(mol K).
Combines Boyle (P ∝ 1/V at const T), Charles (V ∝ T at const P), Avogadro (V ∝ n).
JEE Main Kinetic Theory of Gases — practice questions
Explore 32 free Kinetic Theory of Gases questions in JEE Main Physics. Open each question or exercise to read its content, reference answers and available explanation.
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For an ideal gas, pressure comes from molecular motion and absolute temperature sets the mean translational kinetic energy. Distinguish the rms speed from the average velocity, which can be zero for a gas with no bulk motion.
Key ideas and when to use them
- $pV=nRT$.
- $v_{\rm rms}=\sqrt{3RT/M}$, with molar mass M in kg/mol.
- $p=\tfrac13\rho v_{\rm rms}^2$ for the ideal-gas model.
Common mistakes
- Use temperature in kelvin, not degrees Celsius.
- A temperature ratio changes rms speed by its square root.
Three worked examples
Example 1: Find pV for 2 mol of ideal gas at 300 K. Use R = 8.31 J/(mol·K).
- Apply the ideal-gas equation.
- $pV=2\times8.31\times300$.
4986 J
Example 2: A gas has rms speed 300 m/s at 300 K. Find its rms speed at 1200 K for the same gas.
- The molar mass is unchanged.
- $v_2=300\sqrt{1200/300}$.
600 m/s
Example 3: An ideal gas has density 0.60 kg/m³ and rms molecular speed 100 m/s. Find its pressure.
- Use the rms-speed pressure relation.
- $p=(1/3)\times0.60\times100^2$.
2000 Pa
Qallery AI-assisted study notes and original examples. Further reading: NCERT Physics — Kinetic Theory.
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Practise Kinetic Theory of Gases →Kinetic Theory of Gases questions and exercises with answers
Showing questions 1–30 of 32.
Ideal gas equation:Read the answer and worked solution →Postulate of kinetic theory: gas molecules:Read the answer and worked solution →Avogadro's number:Read the answer and worked solution →R / N_A =Read the answer and worked solution →Root-mean-square speed of gas molecules (mass m):Read the answer and worked solution →Average kinetic energy per molecule (monatomic gas):Read the answer and worked solution →Degrees of freedom for monatomic gas (e.g., He, Ar):Read the answer and worked solution →Most probable speed v_mp:Read the answer and worked solution →Average speed v_avg of molecules:Read the answer and worked solution →Maxwell speed distribution is:Read the answer and worked solution →Internal energy of n moles of ideal gas (monatomic):Read the answer and worked solution →Ratio of specific heats γ for monatomic gas:Read the answer and worked solution →For ideal gas, C_p - C_v =Read the answer and worked solution →Two gases of molar masses M₁ and M₂ at same T. Ratio of v_rms:Read the answer and worked solution →Find v_rms of O₂ at 300 K (M = 32 g/mol, R = 8.314):Read the answer and worked solution →Graham's law of effusion: rate of effusion ∝Read the answer and worked solution →For Maxwell-Boltzmann distribution, fraction of molecules with KE > E behaves as:Read the answer and worked solution →Boyle's law derivation from KTG: at constant T, doubling V causes:Read the answer and worked solution →For monatomic ideal gas, U/T (heat capacity at const V) per mole:Read the answer and worked solution →In adiabatic process for ideal gas:Read the answer and worked solution →For mixture of two ideal gases, total pressure = (Dalton's law):Read the answer and worked solution →Equipartition: at temperature T, each quadratic DOF carries energy:Read the answer and worked solution →Mean free path doubles when:Read the answer and worked solution →At what temperature does v_rms of H₂ equal v_rms of O₂ at 300 K?Read the answer and worked solution →For diatomic gas at very high T (vibrational modes active), DOF = ?Read the answer and worked solution →Boltzmann's distribution gives fraction of molecules in energy state E:Read the answer and worked solution →During adiabatic compression of ideal gas:Read the answer and worked solution →For an ideal gas in adiabatic process: TV^(γ-1) =Read the answer and worked solution →Kinetic energy per molecule of monatomic gas at 300 K (k = 1.38 × 10⁻²³):Read the answer and worked solution →Two containers, equal volume, one with H₂ and one with O₂ at same T, P. Ratio of:Read the answer and worked solution →
Worked practice questions
Postulate of kinetic theory: gas molecules:
- Have fixed positions
- Are in random rapid motion; collisions are elastic
- Are stationary
- Stick together
Show answer and solution
Correct answer: B. Are in random rapid motion; collisions are elastic
KTG postulates: (i) large number of identical molecules in random motion
(ii) elastic collisions (KE conserved)
(iii) negligible molecular volume
(iv) no inter-molecular forces (except during brief collisions).
Avogadro's number:
- 6.022 × 10²³ /mol (number of particles per mole)
- 10⁶
- 13.6
- 9.8
Show answer and solution
Correct answer: A. 6.022 × 10²³ /mol (number of particles per mole)
N_A = 6.022 × 10²³ /mol.
1 mole = N_A elementary entities (atoms, molecules, ions).
Foundation of stoichiometry; relates micro to macro.
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